A train ticket has a perimeter of $36\,\text{centimeters}$ and an area of $80\,\text{square centimeters}$ . What are the dimensions of the ticket? $\newline$ $\_\_\_\_$ centimeters by $\_\_\_\_$ centimeters

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Area and perimeter: word problems

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Q. A train ticket has a perimeter of

36\,\text{centimeters}

and an area of

80\,\text{square centimeters}

. What are the dimensions of the ticket?

\newline

\_\_\_\_

centimeters by

\_\_\_\_

centimeters

Perimeter Equation: Let's denote the length of the ticket as $l$ and the width as $w$ . The perimeter ( $P$ ) of a rectangle is given by the formula $P = 2l + 2w$ . Since we know the perimeter is $36$ centimeters, we can write the equation: $\newline$ $36 = 2l + 2w$
Simplify Perimeter: We can simplify the perimeter equation by dividing all terms by $2$ , which gives us: $\newline$ $18 = l + w$
Area Equation: The area $A$ of a rectangle is given by the formula $A = lw$ . We know the area is $80$ square centimeters, so we can write the equation: $80 = lw$
System of Equations: Now we have a system of two equations with two variables: $\newline$ $1$ . $18 = l + w$ $\newline$ $2$ . $80 = lw$ $\newline$ We can solve this system by expressing one variable in terms of the other using the first equation and then substituting it into the second equation. Let's express $w$ in terms of $l$ : $\newline$ $w = 18 - l$
Quadratic Equation: Substitute $w = 18 - l$ into the area equation $80 = lw$ :
$80 = l(18 - l)$
This simplifies to a quadratic equation:
$80 = 18l - l^2$
Moving all terms to one side gives us:
$l^2 - 18l + 80 = 0$
Solve Quadratic: We need to solve the quadratic equation $l^2 - 18l + 80 = 0$ . This can be factored into: $\newline$ $(l - 10)(l - 8) = 0$ $\newline$ Setting each factor equal to zero gives us two possible solutions for $l$ : $\newline$ $l - 10 = 0$ or $l - 8 = 0$ $\newline$ So, $l = 10$ or $l = 8$
Check Solutions: If $l = 10$ , then substituting back into $w = 18 - l$ gives us $w = 18 - 10$ , so $w = 8$ . If $l = 8$ , then substituting back into $w = 18 - l$ gives us $w = 18 - 8$ , so $w = 10$ . Since a rectangle's length and width are interchangeable, the dimensions of the ticket can be either $10$ centimeters by $8$ centimeters or $8$ centimeters by $10$ centimeters.